A negative charge of -0.0005 C exerts an attractive force of 19.0 N on a second charge that is 25 m
away. What is the magnitude of the second charge?

0.00264C

Explanation:

Using the Columbs law formula

F = kq1q2/r²

F is the force of attraction

q1 and q2 are the charges

r is the distance between the charges.

k is the Coulomb's cinsrsnty

Given F = 19N, q1 = -0.0005C

r = 25m q2 = ?

Substituting

19 = 9×10^9×-0.0005q2/25²

11875 = -4,500,000q2

q2 = 11,875/-4,500,000

q2 = - 0.0264C

Magnitude of the second charge is 0.00264C

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